Answer:
![T_A_v_g=9.918192559$^(\circ)C](https://img.qammunity.org/2021/formulas/physics/college/5tlzlxseq5ysdyzuoeohpppj9zv7k69cny.png)
Step-by-step explanation:
The problem tell us that the temperature as function of time in downtown mathville is given by:
![T(t)=10-5*sin((\pi)/(12t))](https://img.qammunity.org/2021/formulas/physics/college/ifq9aghw3y06motpptv9z78gkf3bgvebdr.png)
The average temperature over a given interval can be calculated as:
![T_a_v_g=(T_o+T_f)/(2)](https://img.qammunity.org/2021/formulas/physics/college/ypyc6mncbt6abvpwr1qmetl8dlt3ctpwxn.png)
Where:
![T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature](https://img.qammunity.org/2021/formulas/physics/college/x9htbop03jaxvxhc4aml3732mpa6uytu20.png)
So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:
Therefore:
![T_o=T(12)=10-5*sin((\pi)/(12*12)) =9.890925575$^(\circ)C](https://img.qammunity.org/2021/formulas/physics/college/ub37tszntp7byzwf8ecqws813ty4ojjcy9.png)
![T_f=T(24)=10-5*sin((\pi)/(12*24)) =9.945459543$^(\circ)C](https://img.qammunity.org/2021/formulas/physics/college/9u9thayshmr8g9s08xsyzigz1q7jcezwh0.png)
Hence, the average temperature between noon and midnight is:
![T_A_v_g=(9.890925575+9.945459543)/(2)=9.918192559$^(\circ)C](https://img.qammunity.org/2021/formulas/physics/college/sndgplfi7vrqlepewmj98owdo8w5lxpmgq.png)